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CircuiTikZ: transformer, okay, but how can I link it correctly?

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Merciadri Luca

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Apr 30, 2010, 12:36:36 PM4/30/10
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Hello,

I am using rhe following code

==
\begin{circuitikz} \draw
(0,0) node[transformer] (T) {}
(T.A1) node[anchor=east] {}
(T.A2) node[anchor=east] {}
(T.B1) node[anchor=west] {}
(T.B2) node[anchor=west] {}
(T.base) node{T}
(T.B2) to[pD] (2,-2)
(T.B1) to[pD] (3,0)
(0.5,-1) to[R=$R$] (3,-1)
(2,-2) to[short, -] (3,-2)
(3,-2) to[short, -] (3,0);
\end{circuitikz}
==

to draw a transformer with median outlet. The problem is that I do not
have the precise coordinates of T.A1, T.A2, T.A3 and T.A4. How can I
have them, or at least draw a straight line (at the bottom) at the
place of a little slopped line that this code draws?

Thanks.
- --
Merciadri Luca
See http://www.student.montefiore.ulg.ac.be/~merciadri/
- --

We tend to be perfect. That’s why when we make mistakes we are hard
on ourselves.
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Merciadri Luca

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Apr 30, 2010, 3:11:51 PM4/30/10
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Merciadri Luca <Luca.Me...@student.ulg.ac.be> writes:

> Hello,
>
> I am using rhe following code
>
> ==
> \begin{circuitikz} \draw
> (0,0) node[transformer] (T) {}
> (T.A1) node[anchor=east] {}
> (T.A2) node[anchor=east] {}
> (T.B1) node[anchor=west] {}
> (T.B2) node[anchor=west] {}
> (T.base) node{T}
> (T.B2) to[pD] (2,-2)
> (T.B1) to[pD] (3,0)
> (0.5,-1) to[R=$R$] (3,-1)
> (2,-2) to[short, -] (3,-2)
> (3,-2) to[short, -] (3,0);
> \end{circuitikz}
> ==
>
> to draw a transformer with median outlet. The problem is that I do not
> have the precise coordinates of T.A1, T.A2, T.A3 and T.A4. How can I
> have them, or at least draw a straight line (at the bottom) at the
> place of a little slopped line that this code draws?

The solution was given to me by the package's author:

==
\begin{circuitikz} \draw
(0,0) node[transformer] (T) {}

(0.5,-1) to[R=$R$] (3.5,-1)
(T.B2) to[pD] ($(T.B2)+(2,0)$) -| (3.5, -1)
(T.B1) to[pD] ($(T.B1)+(2,0)$) -| (3.5, -1)
;
\end{circuitikz}
==

- --
Merciadri Luca
See http://www.student.montefiore.ulg.ac.be/~merciadri/
- --

Where there's a will, there's a way.


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